AMC 12A 2023 (Problem 23)How many ordered pairs of positive real numbers (a,b)(a,b)(a,b) satisfy the equation (1+2a)(2+2b)(2a+b)=32ab(1+2a)(2+2b)(2a+b)=32ab(1+2a)(2+2b)(2a+b)=32ab?(A) 0\text{(A)}\;0(A)0(B) 1\text{(B)}\;1(B)1(C) 2\text{(C)}\;2(C)2(D) 3\text{(D)}\;3(D)3(E) an infinite number\text{(E)}\;\text{an infinite number}(E)an infinite numberRelated TopicsToolkit 28 — Max-QM-AM-GM-HM-Min InequalityHints (7)Hint 1Use 28. AM–GM Inequality.Hint 21+2a ≥ 22a1+2a\ \ge\ 2\sqrt{2a}1+2a ≥ 22aHint 32+2b ≥ 24b=4b2+2b\ \ge\ 2\sqrt{4b}=4\sqrt{b}2+2b ≥ 24b=4bHint 42a+b ≥ 22ab2a+b\ \ge\ 2\sqrt{2ab}2a+b ≥ 22abHint 5By multiplying Hints 2, 3, and 4: (1+2a)(2+2b)(2a+b) ≥ 32ab(1+2a)(2+2b)(2a+b)\ \ge\ 32ab(1+2a)(2+2b)(2a+b) ≥ 32abHint 6By using 28. AM–GM Inequality, equality holds if and only if 1=2a, 2=2b, 2a=b1=2a,\ 2=2b,\ 2a=b1=2a, 2=2b, 2a=bHint 7a=12,b=1a=\tfrac12,\quad b=1a=21,b=1Final Answer(B) 1